abc numar =3abnumar+2bc numar +ca numar
100a +10b+c=3(10a+b)+2(10b+c)+10c+a
100a+10b+c=30a+3b+20b+2c+10c+a
100a+c=31a+13b+12c
69a=13b+11c
1≤a,b,c≤9
pta=1
13b+11c=69
b=(69-11c):13
ptc=1, b=(69-11):13=58:13∉N
c=2 , b= 47:13∉N
c=3, b=36:13∉N
c=4 b=25:13∉N
c=5 b=14:13∉N
pt c=6,7,8,9 nu mai incercam, b devine<1 si apoi <0
pt a=2
13b+11c=69*2=138
b=(138-11c):13
pt c=1 b=127:13∉N
c=2 b=116:13∉N
c=3 b=105:13∉N
c=4 b=94:13∉N
c=5 b=83:13∉N
c=6 b=72:13∉N
c=7 b=61:13∉N
c=8 b=50:13∉N
c=9 b=39:13=3∈N SOLUTIE BUNA
Deci
a=2, b=3, c=9
Verificare
239=3*23+2*39+92
239=69+68+92
239=69+170
239=239
Adevarat
pt a=3,
13b+11c=69*3=207
b=(207-11c):13
pt c∈{1;2;...9}
b∈{196/13;185/13;174/13;163/13;152/13;141/13; 130/13; 119/13;108/13} ∩N =10
dar 10>9, nu convine, nu este cifra
deci pt a=3 nu avem solutii
pt a=4
13b+11c=69*4=276
b= (276-11c):13∈{265/13;...177/13} dintre care oricare este>10, (nu mai e nevoie sa verificam si daca∈N) deci nu poate fi cifra
la fel pt oricare a≥5, si 1≤c≤9, va rezulta b>10 , deci NU mai incercam pt valori ale lui a =5,6,7,8,9
DEci am explorat TOATE posibilitatile, si SINGURA solutie este
a=2 ;c=3; b=9
